Question

Integration Using Trigonometric Identities 

Answer 1

In this section, we use known trigonometric identities to find the integral. When the integrand involves some trigonometric functions, we use the following trigonometric identities:

• sin²x = \(\frac{1-cos2x}{2}\)
• cos²x = \(\frac{1+cos2x}{2}\)
• sin (A + B) + sin (A - B) = 2 sin A cos B
• sin (A + B) - sin (A - B) = 2 cos A sin B
• cos (A + B) + cos (A - B) = 2 cos A cos B
• cos (A - B) - cos (A + B) = 2 sin A sin B
• sin³x = \(\frac{3\ sin\ x \ - sin\ 3x}{4}\)
• cos³x = \(\frac{cos\ 3x \ + 3\ cos\ x}{4}\)

Integrals of some Particular Functions

In this section, we introduce some important formulae of integrals and apply them to evaluate many other integrals:

Expression	Substitution
(A) x2 + a2	x = a tan θ or a cot θ
(B) x2 - a2	x = a sec θ or a cosec θ
(C) a2 - x2	x = a sin θ or a cos θ
(D) \(\sqrt{\frac{a-x}{a+x}} \ or \ \sqrt{\frac{a+x}{a-x}} \)	x = a cos 2θ

Theorem:

(i) \(\int \frac{1}{x^2 + a^2}\) dx = \(\frac{1}{a} \) tan^-1 \(\frac{x}{a}\) + C

Proof:

We now prove the above results.

Putting x = a tan θ

Then dx = a sec²θ

and tan θ = x/a

or θ = tan^-1x/a

(ii) \(\int \frac{1}{x^2 -a^2} dx = \frac{1}{2a} log |\frac{x-a}{x+a}| + C\)

Proof:

Alter: I = \(\int \frac{1}{x^2 - a^2} \ dx\)

Putting x = a sec θ

Then dx = a sec θ tan θ dθ

and sec θ = \(\frac{x}{a}\)

θ = sec^-1 \(\frac{x}{a}\)

(iii) \(\int \frac{1}{a^2 - x^2}\ dx = \frac{1}{2a} log |\frac{a+x}{a-x}| + C\)

Proof:

(iv) \(\int \frac{1}{\sqrt{a^2 - x^2}} \) dx = sin^-1 \((\frac{x}{a}) + C\)

Proof:

Let I = \(\int \frac{1}{\sqrt{a^2 - x^2}} \) dx

Putting x = a sin θ

Then dx = a cos θ dθ

and sin θ = \(\frac{x}{a}\)

or θ = sin^-1 \(\frac{x}{a}\)

Similarly, by making substitution x = a cos θ, we get

\(\int \frac{1}{\sqrt{a^2 -x^2}} dx = -cos^{-1} \frac{x}{a} + C\)

(v) \(\int \frac{}{\sqrt{a^2 + x^2}}\ dx = log |x + \sqrt{a^2 + x^2}| + C\)

Proof:

Let I = \(\int \frac{1}{\sqrt{a^2 +x^2}} dx\)

Putting x = a tan θ

Then dx = a sec² θ dθ

and tan θ = \(\frac{x}{a}\)

or θ = tan^-1 \(\frac{x}{a}\)

(vi) \(\int \frac{1}{\sqrt{x^2 - a^2}}\ dx = log |x + \sqrt{x^2 - a^2}| + C\)

Proof:

Let I = \(\int \frac{1}{\sqrt{x^2 - a^2}}\ dx\)

Putting x = atan θ

Then dx = a sec θ tan θ dθ

and tan θ = \(\frac{x}{a}\)

or θ = tan^-1 \(\frac{x}{a}\)